Answer:
t= 27.38 mins [this the time taken by the enzyme to hydrolyse 80% of the fat present]
Explanation:given values
Half life of lipase t_1/2 = 8 min x 60s/min = 480 s
Rate constant for first order reaction
k_d = 0.6932/480 = 1.44 x 10^-3 s-1
Initial fat concentration S_0 = 45 mol/m3 = 45 mmol/L
rate of hydrolysis Vm0 = 0.07 mmol/L/s
Conversion X = 0.80
Final concentration S = S_0(1-X) = 45 (1-0.80) = 9 mol/m3
K_m = 5mmol/L
time take is given by
![t= -\frac{1}{K_d}ln[1-\frac{K_d}{V_m_0}(k_mln\frac{s_0}{s}+(s_0-s))]](https://tex.z-dn.net/?f=t%3D%20-%5Cfrac%7B1%7D%7BK_d%7Dln%5B1-%5Cfrac%7BK_d%7D%7BV_m_0%7D%28k_mln%5Cfrac%7Bs_0%7D%7Bs%7D%2B%28s_0-s%29%29%5D)
all values are given and putting these value we get
t=1642.83 secs
which is equal to
t= 27.38 mins [this the time taken by the enzyme to hydrolyse 80% of the fat present]
3.39 x 10^-13
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The pain was triggered by the Nervous System
Answer:
a) 27.2 V
b)27.2 V
Explanation:
Charge of the electron =charge of the proton = q = 1.6 × 10⁻¹⁹ C
Radius = r = 0.53×10⁻¹⁰ m
Electric Potential = V = k q/r
k = 9 ×10⁹ N m²/C² = Coulomb's constant.
V = (9 ×10⁹)(1.6 × 10⁻¹⁹)/( 0.53×10⁻¹⁰) = 27.2 V
b) Potential Energy of the electron = k q × q / r
= [(9 ×10⁹)(1.6 × 10⁻¹⁹)(1.6 × 10⁻¹⁹) / (0.53×10⁻¹⁰)] / (1.6 × 10⁻¹⁹) eV,
since 1 electron volt = (1.6 × 10⁻¹⁹)joules
= 27.2 eV